EDEXCEL NATIONAL CERTIFICATE/DIPLOMA …

1 1 EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME 2 ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS 2 ENGINEERING COMPONENTS Structural members: struts and ties; direct stress and strain, dimensional changes; combined effects of direct and thermal loading, factor of safety. Compound members: series and parallel connected compound bars made up of two materials; direct stress and strain in each material, dimensional changes. Fastenings: shear stress in fastenings riveted joints, bolted joints and hinge pins subjected to single and double shearing forces You should judge your progress by completing the self assessment exercises.

2 These may be sent for marking at a cost (see home page). On completion of this tutorial you should be able to do the following. Define structural members. Calculate direct stress and strain. Calculate changes in dimensions. Solve basic problems involving stress, strain and modulus. Explain and calculate Safety Factor. Explain and calculate stresses due to temperature changes. It is assumed that the student is already familiar with the concepts of FORCE. 1. TYPES OF STRUCTURAL MEMBERS Engineering structures come in many forms. Here are some. STRUTS AND TIES A strut is a long thin member that is compressed and usually fails by buckling.

3 A tie is a member that is stretched so it cannot buckle. A tie could be a rope or chain as well as a rigid length of material. Figure 1 FRAMES Struts and ties make up the members of lattice frames such as the simple one shown. The two side members are compressed and so are struts but the bottom one is stretched and could be a chain so it is a tie. Figure 2 COLUMNS A column is a thick compression member. Struts fail due to bending but columns fail in compression. Columns are usually made of brittle material which is strong in compression such as cast iron, stone and concrete. These materials are weak in tension so it is important to ensure that bending does not produce tensile stresses in them.

4 If the compressive stress is too big, they fail by crumbling and cracking Figure 3 2 BEAMS A beam is a structure, which is loaded transversely (sideways). The loads may be point loads or uniformly distributed loads (udl). The diagrams show the way that point loads and uniform loads are illustrated. Figure 4 Transverse loading causes bending and bending is a very severe form of stressing a structure. The bent beam goes into tension (stretched) on one side and compression on the other. Figure 5 2. DIRECT STRESS When a force is applied to an elastic body, the body deforms.

5 The way in which the body deforms depends upon the type of force applied to it. A compression force makes the body shorter. A tensile force makes the body longer. Figure 6 Tensile and compressive forces are called DIRECT FORCES. Stress is the force per unit area upon which it acts. Stress = = Force/Area N/m2 or Pascals. The symbol is called SIGMA NOTE ON UNITS The fundamental unit of stress is 1 N/m2 and this is called a Pascal. This is a small quantity in most fields of engineering so we use the multiples kPa, MPa and GPa. Areas may be calculated in mm2 and units of stress in N/mm2 are quite acceptable. Since 1 N/mm2 converts to 1 000 000 N/m2 then it follows that the N/mm2 is the same as a MPa 3 3.

6 DIRECT STRAIN In each case, a force F produces a deformation x. In engineering we usually change this force into stress and the deformation into strain and we define these as follows. Strain is the deformation per unit of the original length Strain = = x/L The symbol is called EPSILON Strain has no units since it is a ratio of length to length. Most engineering materials do not stretch very much before they become damaged so strain values are very small figures. It is quite normal to change small numbers in to the exponent for of 10-6. Engineers use the abbreviation (micro strain) to denote this multiple. For example a strain of could be written as 68 x 10-6 but engineers would write 68.

7 Note that when conducting a British Standard tensile test the symbols for original area are So and for Length is Lo. WORKED EXAMPLE A metal wire is mm diameter and 2 m long. A force of 12 N is applied to it and it stretches mm. Assume the material is elastic. Determine the following. i. The stress in the wire . ii. The strain in the wire . SOLUTION 2222N/mm AF mm x 4 dA====== Answer (i) is hence MPa 150or Lx=== SELF ASSESSMENT EXERCISE 1. A steel bar is 10 mm diameter and 2 m long. It is stretched with a force of 20 kN and extends by mm. Calculate the stress and strain. (Answers MPa and 100 ) 2. A rod is m long and 5 mm diameter.

8 It is stretched mm by a force of 3 kN. Calculate the stress and strain. (Answers MPa and 120 ) 4 4. MODULUS OF ELASTICITY E Elastic materials always spring back into shape when released. They also obey HOOKE'S LAW. This is the law of a spring which states that deformation is directly proportional to the force. F/x = stiffness = k N/m Figure 7 The stiffness is different for different materials and different sizes of the material. We may eliminate the size by using stress and strain instead of force and deformation as follows. If F and x refer to direct stress and strain then F = A x = L hence ==AxFL and L AxF The stiffness is now in terms of stress and strain only and this constant is called the MODULUS of ELASTICITY and it has a symbol E.

9 ==AxFL E A graph of stress against strain will be a straight line with a gradient of E. The units of E are the same as the units of stress. 6. ULTIMATE TENSILE STRESS If a material is stretched until it breaks, the tensile stress has reached the absolute limit and this stress level is called the ultimate tensile stress. Values for different materials may be found in various sources such as the web site Matweb. WORKED EXAMPLE A steel tensile test specimen has a cross sectional area of 100 mm2 and a gauge length of 50 mm, the gradient of the elastic section is 410 x 103 N/mm.

10 Determine the modulus of elasticity. SOLUTION The gradient gives the ratio F/A = and this may be used to find E. GPa 205or MPa 000 205or N/mm 000 205 100 50 x 10 x 410 ALxxF E23==== 5 WORKED EXAMPLE A Steel column is 3 m long and m diameter. It carries a load of 50 MN. Given that the modulus of elasticity is 200 GPa, calculate the compressive stress and strain and determine how much the column is compressed. SOLUTION mm mm 3000 x xso Lx x10200 so EPa x1050AF m x 4 dA9666222============== 5. SAFETY FACTOR The stress at which a material is deemed to fail might be the ultimate stress or the yield stress.